Question

find the sum of first 8 multiples of 3

Answer 1

From Arithmetic Progresses, sum of n terms from 1st term is $\dfrac{n}{2}[2a+(n-1)d]$


In the given question,

first term = first multiple of 3
first term = 3

common difference = 3




Then, applying formula from above.


$S_{n}=\dfrac{n}{2}[2a+(n-1)d]$


Assuming the value of n as 8 and a as 3


$S_{8} =\dfrac{8}{2}[2(3)+(8-1)3]$


$S_{8}=4[6+7(3)]$


$S_{8}=4[6+21]$


$S_{8}=4[27]$


$S_{8}=108$





Therefore sum of first 8 multiples of 3 is 108.

Answer 2

First 8 multiples of 3 = 3,6,9,12,15,18,21,24
Their sum = 3 + 6 + 9 + 12 + 15 + 18 + 21 + 24
= 108 Answer
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